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Unformatted text preview: Math 110, HW12 Avik Das, SID: 20297989
November 23, 2011 Section 6.6 1 0 1. (a) False. Consider the projection T on span {(1, 1)} along (0,1) in R2. Then, [T] 3 = (1 0 >, which
i is not self—adjoint, so T is not selfadjoint. ‘ _ ‘ , r _ (b) True. Given the range of an orthogonal projection T, we know that T is the projection on R (T)
[If f: r_ _j' 5 :2 'di. U 'A‘._:VJHI,:,'". f ' ;{  alongN(T)—_—R(T)i.
' ' ‘ ' ‘ ' “ " T~T;{ﬂ ‘Vr'fijjr‘ ' Jpqﬁf'VT afgi} (c) True. Spectral Theorem.
‘ l L I V I ‘ (d) False. This is only true when T is an orthogonal projection. (e) False. By Corollary 2 of the Spectral Theorem, this requires that [A] = 1 for every eigenvalue
lambda of the projection. 2. Projection of R2 on span({(1,2)}): T(1,0) = (é, T(0,1) =
( l 2
§,§). Thus, [T], =(% g)
Projection of R3 on span ({(1,0,1)}): T(1,0,0) = ( (0, 1,0) 6 l
2% o %
span({1,0,1})‘l'), T(0,0,1) = (5,03). Thus, [T], = ( (I) 0 (3 , _ 0 _
2 2 5. (a) To show that S it is sufﬁcient to Show that S Let :c 2 ml + 102
where ml and 1.02 are orthogonal and T = wl. Then, = (wl + 1112,1111 + 102) = (101,101) +
(102,101) + (101,102) + (1112402). Because ml and 'wg are orthogonal, (w1,w2) = (102,101) = 0, so , H96“2 =UI12 + Hw2l2 2 Huh”2 = HT($)H2 Consider the projection Ton span{(1,1)} along (0,1). Then, T (—2, —1) = (—2, —2). HT (—2, —1) = “(‘27 2) = \/§ > V5 = “(—2» “1)~
If the inequality is actually an equality for all a: E V, then the projection is simply the identity
in ,.' r. .. r 1 maponV. . : (b) From (a), we have that u was)“ 3 use”, then “qu + <w2,w1> + <wuw2> +uw212 2 “qu2.
' ‘ ‘ 1 meaning (wg,'wl)+(1111,w2)—+—ng]]2 2 0 => ngll2 2 —'(w2,w1)——(w1,w2). Consider x1 : w1+w2 and $2 = w1 —w2. The inequality must hold for both :31 and 902, so “102”2 2 — (w2,w1) — (101,102) and “1.02”2 2 — (—w2,w1) ~ (w1,~w2) = (1112,1121) + (101,102). Thus, it follows that (w2,w1> + (1111,1122) = 0, meaning wl and 1112 are orthogonal. Thus, T is an orthogonal projection. n  ,, j g n I 7. (a) Let g(t) = 29:104937. Then, g(T) = 9 (2:1 AiT¢> = ijlaj(2f=1AiT,) . Note that _~ y  _: r _ _: _ _ __. . g If _ .V, because T: = T, for any j 2 1, and TaTl7 = To for 1 3 a,b g k, so (21;, AiTi) = Zf=1A§T,.
. _ _ _ _ . _ . . H _ ._ Thus, 90—) 2 23,121,” 2;; )‘gT, = 2le 239:, ajAgT, = 2:13.23le ajAg' 2 2:13.90”) (b) Let g (t) = t". Then, g(T) = 217:1 M‘T, = To. Because each T, is an orthogonal projection, it
follows that A? = 0 for 1 g i g k, meaning A, = 0. Thus, T = 21;, m, = 21;, T0 = To. (c) UT = TU iff 217:1 UT, 2 2L1 T,~U iﬁ U commutes with each Ti. (01) Let U = 21;, VET, Let g(t) = 132. Then, U2: g(U) = 21;, 9 (ME) T, = 21;, Am. (e) T is invertible iff rank (T) = dim (V), meaning for any x 0, T aé 0 O Thus, none of the
eigenvalues of T are 0. The A, are exactly the distinct eigenvalues of T, so T is invertible iff all of
the eigenvalues of T are nonzero iii A, # 0 for 1 S i S k. (f) T is a projection iff T2 = T iff T2 = 2L1 AfTi = 217:1 AiT, = T iff A? = A, for 1 S i g k iff
A, = 1 or A, = 0 for 1 S i S k. Note that the A, are the distinct eigenvalues of T, T is a projection
iff every eigenvalue of T is 1 or 0. (g) Note that each T,» is an orthogonal projection, so T: = T,. T* 2 (2le A,T,) =2:le A_,T;k = Zf=1A_,T,. Thus, T = —T* iff A, = aALfor 1 S i S k. This means that A, is an imaginary
number (since if A, = a + bi = —a + bi : A1, then a = —a 3 a = 0). Section 6.7 " 1. (a) False. Consider the case where the operator, T, is self—adjoint. Then, given a singular value 0', it
may be that —a is an eigenvalue of T and not a (see part (e)). (b) False. The eigenvalues of A*A are the squares of the sigunal values. (c) True. Note that if A is an eigenvalue for A, then AA is an eigenvalue for A*A, and a = VAA is
a singular value of A. Furthermore, cA is an eigenvalue of CA, so JCA == ECAA = c2AA is an eigenvalue for (CA)* (CA). Thus, V c2AA 2 {cl VAA = c a is a singular value for 0A., ((1) True. Singular Value Theorem for Linear Transformations states that the singular value a, is
positive for 1 S i S 7‘ and the deﬁnition of singular value states that 0‘, = 0 for i 2 7", where r is
the rank of the operator. \/ (e) False. This is only true if A > 0, otherwise the corresponding singular value is —A.
(f) False. This is only true if b E R ya
(g) True. The pseudoinverse is deﬁned thatfzay. 1 0 .
2. (a) A = (I 1 >, so A*A = An orthonormal basis for R2 consisting of eigenvectors
1 —1 of A*A is , = {111,122} with corresponding eigenvalues 3, 2. Then, an orthonormal 1 O subset of R3 is {ﬁg = 715 ,x/iiLA = f < 11) This can be extended to
’ l ' , 1 0 —2 an orthonormal basis for R3: % i ' ,ﬁ 11 q}? i : {U1,U2,U3}. Finally, let 01=\/§,02=\/§ 0 0 2 0 0
(b) A = (0 0) , so A*A = ( 0 0) .' An orthonormal basis for P2 consisting of eigenvectors
0 0 4 (E) (13) v (i) 0
orthonormal subset of P2 is {%LA (0) = (1) This can be extended to an orthonormal basis 0
0
0
} = {111,112,123} with corresponding eigenvalues 4,0,0. Then, an 1 0 for P2: , = {uhuQ}. Finally, let 01 = 2. 13. 26. l)‘ i 1 If A 0 = amtvi 2 Jim. Thus, by construction, UZV* is an SVD for A.
2' ' 'W l 2' = A'Ui = Aim = { Because A is positive semideﬁnite, it is Hermitian and all of its eigenvalues are nonnegative. Let
A1, . . . , An be the eigenvalues of A. Then, the eigenvalues of A*A = A2 are simply A? for i = 1, . . . , n. Then, the singular values of A are MAE, but because the M are nonnegative, the singular values are
simply Ai. Letﬁ: v1,...,vn and7= u1,...,vum . LetA= T7 and let on be the singular values ofAfor
[3 31w iflgigr 0 yifr<i§m =[LA+1€= AT = (my. 1': 1,. .. ,7", where r 2 rank Then LAT (11.1) = (LA)Jr ={ . Note that LA 2 T, so (LA)Jf = TJr (ui) = . Thus, [Tl]:i = [(LA)T]B ,y 6;”); iflgigr
0 ifr<igm ...
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 Fall '07
 Hutchings

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